Question

The value of $\Delta {H_f}$ of $N{O_{(g)}}$ is?
$\
  {N_{2(g)}} + 2{O_{2(g)}} \to 2N{O_{2(g)}} + xkJ \\
  2N{O_{(g)}} + {O_{2(g)}} \to 2N{O_{2(g)}} + ykJ \\
\ $
A. $\dfrac{1}{2}(y - x)$
B. $\dfrac{1}{2}(x - y)$
C. $(x - y)$
D. $2(x - y)$

Answer 1

Hint:$\Delta {H_f}$ represents the heat of formation, it is calculated by the difference between the enthalpy of the product minus the enthalpy of the reactants. It is measured in kilojoule per mole that is $kJ/mol$ . More precisely, it is the change in enthalpy for creating one mole of a compound at standard conditions.

Complete step by step answer:
According to the question:
There are two chemical equation:
${N_{2(g)}} + 2{O_{2(g)}} \to 2N{O_{2(g)}} + xkJ$ (i)
$2N{O_{(g)}} + {O_{2(g)}} \to 2N{O_{2(g)}} + ykJ$ (ii)
And we have to calculate the $\Delta {H_f}$ of $N{O_{(g)}}$ .
The reaction for $N{O_{(g)}}$ is: $\dfrac{1}{2}{N_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to N{O_{(g)}}$
In order to determine its enthalpy of formation we have to reverse the second given equation in the question. That becomes,
 $2N{O_{2(g)}} + ykJ \to 2N{O_{(g)}} + {O_{2(g)}}$ (iii)
Let’s name that reaction the third reaction.
Now, in order to form a third equation by the given equations we have to add equation (i) and (iii). That becomes,
${N_{2(g)}} + 2{O_{2(g)}} \to 2N{O_{2(g)}} + xkJ$ (i)
$2N{O_{2(g)}} + ykJ \to 2N{O_{(g)}} + {O_{2(g)}}$ (iii)
On addition they form a new equation as:
\[{N_{2(g)}} + 2{O_{2(g)}} + 2N{O_{2(g)}} + ykJ \to 2N{O_{2(g)}} + 2N{O_{(g)}} + {O_{2(g)}} + xkJ\]
On solving this equation becomes:
${N_{2(g)}} + {O_{2(g)}} + ykJ \to 2N{O_{(g)}} + xkJ$
On simplifying:
${N_{2(g)}} + {O_{2(g)}} \to 2N{O_{(g)}} + xkJ - ykJ$
That is:
$\dfrac{1}{2}{N_{2(g)}} + \dfrac{1}{2}{O_{2(g)}} \to N{O_{(g)}} + \dfrac{1}{2}(x - y)kJ$
As it is a endothermic reaction so its enthalpy of formation will be negative:
So, enthalpy of formation of $N{O_{(g)}}$ will be $ - \dfrac{1}{2}(x - y)kJ/mol $
$ \Rightarrow \dfrac{1}{2}(y - x)kJ/mol $
So, $\Delta {H_f} = \dfrac{1}{2}(y - x)$

Hence, option A is correct.

Note:
Sometimes the formation of reaction may be defined as the sum of a number of simpler reactions, does not matter if they are real or fictitious. The enthalpy of reaction can then be determined by the help of Hess's Law, this law states that the sum of the enthalpy changes for a number of individual reaction steps equals to the enthalpy change of the overall reaction. This is true because enthalpy is a state function, in state function value for an overall process depends only on the initial and final states and not on any intermediate states.